SOLUTION: Find the largest positive integer $n$ such that \[\frac{(n + 1)^2}{n + 2}\] is an integer.

Algebra ->  Divisibility and Prime Numbers -> SOLUTION: Find the largest positive integer $n$ such that \[\frac{(n + 1)^2}{n + 2}\] is an integer.      Log On


   

Question 1207642: Find the largest positive integer $n$ such that
\[\frac{(n + 1)^2}{n + 2}\]
is an integer.
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

I wish you wouldn't use notation incompatible with HTML.

-------------------------------------------------------

Find the largest positive integer n such that

%28n%2B1%29%5E2%2F%28n%2B2%29 is an integer?

I don't think it ever can be a positive integer if n is a positive integer.
Let's see why:

%28n%5E2%2B2n%2B1%29%2F%28n%2B2%5E%22%22%29

Divide it out by long division:

             n + 0
n + 2) n2 + 2n + 1
       n2 + 2n
            0n + 1
            0n + 0
                 1   
       
%28n%5E2%2B2n%2B1%29%2F%28n%2B2%29%22%22=%22%22n%2B1%2F%28n%2B2%29 and thus will never be a positive 
integer, because 1%2F%28n%2B2%29 will always be a fraction and will
never be a positive integer to add to positive integer n to 
give a positive integer.

Edwin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

As tutor Edwin points out in his long division table, the 1 at the very bottom represents remainder 1.

This means that computing %28%28n%2B1%29%5E2%29%2F%28n%2B2%29 will give some quotient with remainder 1.

Here are some examples:
n = 5 --->
and
n = 6 --->
and
n = 7 --->

Getting remainder 1 means (n+2) is not a factor of (n+1)^2, and furthermore means %28%28n%2B1%29%5E2%29%2F%28n%2B2%29 is never an integer when n is a positive integer.


If we got remainder 0, then it would prove the fraction is an integer.

-------------------------------------------------------------------------

Another approach that doesn't use long division

%28%28n%2B1%29%5E2%29%2F%28n%2B2%29

= %28n%5E2%2B2n%2B1%29%2F%28n%2B2%29

= %28%28n%5E2%2B2n%29%2B1%29%2F%28n%2B2%29

= %28n%5E2%2B2n%29%2F%28n%2B2%29%2B1%2F%28n%2B2%29

= %28n%28n%2B2%29%29%2F%28n%2B2%29%2B1%2F%28n%2B2%29

= n%2B1%2F%28n%2B2%29

In short,
%28%28n%2B1%29%5E2%29%2F%28n%2B2%29+=+n%2B1%2F%28n%2B2%29

We get a whole number n, but an additional fractional amount 1/(n+2) to prove that n%2B1%2F%28n%2B2%29 is never an integer.